If A = $\begin{bmatrix$}[r] 4 & x \ 0 & 1 \end{bmatrix}, B = \begin{bmatrix}[r] 2 & 12 \ 0 & 1 \end{bmatrix} and A = B², the value of x is :

1. 38

2. -6

3. -36

4. 36

Substituting value of A and B in A = B², we get :

$\Rightarrow \begin{bmatrix$}[r] 4 & x \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 2 & 12 \ 0 & 1 \end{bmatrix}\begin{bmatrix}[r] 2 & 12 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & x \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 2 \times 2 + 12 \times 0 & 2 \times 12 + 12 \times 1 \ 0 \times 2 + 1 \times 0 & 0 \times 12 + 1 \times 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & x \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 4 + 0 & 24 + 12 \ 0 + 0 & 0 + 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & x \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 4 & 36 \ 0 & 1 \end{bmatrix}

∴ x = 36.

Hence, Option 4 is the correct option.